HMMT 二月 2004 · 冲刺赛 · 第 17 题

HMMT February 2004 — Guts Round — Problem 17

[8] Kate has four red socks and four blue socks. If she randomly divides these eight socks into four pairs, what is the probability that none of the pairs will be mismatched? That is, what is the probability that each pair will consist either of two red socks or of two blue socks?

解析

Kate has four red socks and four blue socks. If she randomly divides these eight socks into four pairs, what is the probability that none of the pairs will be mismatched? That is, what is the probability that each pair will consist either of two red socks or of two blue socks? Solution: 3 / 35 ( ) 4 The number of ways Kate can divide the four red socks into two pairs is / 2 = 3. The 2 number of ways she can divide the four blue socks into two pairs is also 3. Therefore, the number of ways she can form two pairs of red socks and two pairs of blue socks is 4 3 · 3 = 9. The total number of ways she can divide the eight socks into four pairs is [8! / (2! · 2! · 2! · 2!)] / 4! = 105, so the probability that the socks come out paired correctly is 9 / 105 = 3 / 35. To see why 105 is the correct denominator, we can look at each 2! term as representing the double counting of pair ( ab ) and pair ( ba ), while the 4! term represents the number of different orders in which we can select the same four pairs. Alternatively, we know that there are three ways to select two pairs from four socks. To select three pairs from six socks, there are five different choices for the first sock’s partner and then three ways to pair up the remaining four socks, for a total of 5 · 3 = 15 pairings. To select four pairs from eight socks, there are seven different choices for the first sock’s partner and then fifteen ways to pair up the remaining six socks, for a total of 7 · 15 = 105 pairings.